All categories

3 years ago

The length and width of a rectangle whose length is 11 centimeters more than its width

Mathematics

1 answer:

sammy [17]3 years ago

3 0

Answer:

The length is $x\ cm$ and width is $x+11\ cm$ .

Step-by-step explanation:

Given:

Length is 11 centimeters more than its width.

Now, to find the length and width.

Let the width of rectangle be $x\ cm.$

And the length of rectangle = $x+11\ cm.$

Therefore, the length is $x\ cm$ and width is $x+11\ cm$ of the rectangle.

You might be interested in

Pls help I’ll give you 35 points A book is on sale for

Phantasy [73]

The original price of the book is 16$ the discount is 4$. 4$ is 25% of 16$

8 0

3 years ago

Help me on this question please?

lozanna [386]

The answer to the first part is D
The answer to the second part is A

3 0

4 years ago

If the degree in the numerator is smaller than the degree in the denominator and there is no vertical

mote1985 [20]

The answer is false.

8 0

3 years ago

Read 2 more answers

Show all work to identify the asymptotes and zero of the function f(x) = 6x / x^2 - 36

eduard

Answer:

Zero of the function f(x) is at x = 0

Vertical Asymptotes at x = ±6

Horizontal Asymptotes at y = 0

Step-by-step explanation:

<h3>Vertical Asymptotes </h3>

For a given function f(x):

Vertical Asymptotes are obtained at those values of x, where the function f(x) tends to infinity, I.e.,

When x approaches some constant value but the curve moves towards infinity.

If f(x) is a fraction, it'll tend to infinity when it's denominator becomes zero.

Vertical Asymptotes of the given function can be obtained by walking thru the following steps:

Step I

(Factorise the numerator and denominator)

$\mathsf{ f(x) = \frac{6x}{ {x}^{2} - 36 } }$

x² - 36 can be factorised into (x + 6)(x - 6)

and, ofcourse, we can write 6x as 6(x - 0)

$\mathsf{ f(x) = \frac{6(x - 0)}{ (x + 6)(x - 6) } }$

Step II

(Reduce the fraction to its simplest form by canceling out the common factors)

There aren't any common factors in the numerator and denominator in this case.

Step III

(Look for the values of x which cause the denominator to be zero)

If we put x = 6

denominator becomes 0

Also,

If we substitute x with -6

denominator becomes 0.

The two values of x indicate the two Vertical Asymptotes of the function f(x).

Therefore,

Vertical Asymptotes of the given function f(x) are:

$\boxed{ \mathsf{x = \pm6}}$

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

<h3 /><h3>Horizontal Asymptotes:</h3>

Horizontal Asymptotes are obtained When x tends to infinity and y approaches some constant value.

I'll be using the concept of limits for this.

$\mathsf{y = \frac{6x}{ {x}^{2} - 36 } }$

dividing and multiplying by x² (Yep! so if x becomes infinity 1/ x and 1/ x² all such terms become 0, 'cause 1/ ∞ is 0)

$\implies \mathsf{y = lim_{x \rightarrow \infty }( \frac{ \frac{6x}{ {x}^{2} } }{ \frac{ {x}^{2} - 36 }{ {x}^{2} } } ) }$

$\implies \mathsf{y = lim_{x \rightarrow \infty }( \frac{ \frac{6}{ x } }{ 1- \frac{36 }{ {x}^{2} } } ) }$

Substitute x with ∞, you get zero/ 1

$\implies \boxed{\mathsf{y = 0}}$

So, the horizontal Asymptote of the function is y = 0, that is the x axis

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

<h3>Zeroes of a function:</h3>

The values of x that reduces f(x) to zero are called the zeroes of f(x).

Here, only x = 0 acts as the zero of the function.

[NOTE:

For finding Vertical Asymptotes,Equate the denominator to 0. And
For finding Zeroes, Equate the numerator to 0]

__________________

[That's what it's graph looks like. ]

3 0

3 years ago

Please answer quick. I will mark brainliest

Elena L [17]

Answer:

a=5

ʕ•ᴥ•ʔʕ•ᴥ•ʔʕ•ᴥ•ʔʕ•ᴥ•ʔʕ•ᴥ•ʔʕ•ᴥ•ʔʕ•ᴥ•ʔʕ•ᴥ•ʔ

Take a bear and have a nice day.

5 0

3 years ago